Question

find the sum of all integer between 100and 550,which are divisible by 9

Answer 1

AP : 108,117,126,....549
a = 108 d = 9
Last term = 549

An = 549

a+(n-1)d = 549

108+(n-1)9 = 549

9n-9 = 549-108

9n = 441+9

9n = 450 => n = 50

So, There is 50 numbers between 100 and 550 which are divisible by 9

Now,

Sum of all integer between 100and 550,which are divisible by 9,
i.e.,
Sn = (n/2){2a+(n-1)d}

= (50/2){2×108+(50-1)9}

= 25(216+49×9)

= 25(216+441)

= 25×657

= 16425